# How do you find the equation of the perpendicular bisector of the points (1,4) and (5,-2)?

May 12, 2018

$y = \frac{2}{3} x - 1$

#### Explanation:

$\text{a perpendicular bisector, bisects a line segment at}$
$\text{right angles}$

$\text{to obtain the equation we require slope and a point on it}$

$\text{find the midpoint and slope of the given points}$

$\text{midpoint } = \left[\frac{1}{2} \left(1 + 5\right) , \frac{1}{2} \left(4 - 2\right)\right]$

color(white)("midpoint ")=(3,1)larrcolor(blue)"point on bisector"

$\text{calculate the slope m using the "color(blue)"gradient formula}$

•color(white)(x)m=(y_2-y_1)/(x_2-x_1)

$\text{let "(x_1,y_1)=(1,4)" and } \left({x}_{2} , {y}_{2}\right) = \left(5 , - 2\right)$

$\Rightarrow m = \frac{- 2 - 4}{5 - 1} = \frac{- 6}{4} = - \frac{3}{2}$

$\text{given a line with slope m then the slope of a line}$
$\text{perpendicular to it is}$

•color(white)(x)m_(color(red)"perpendicular")=-1/m

rArrm_("perpendicular")=-1/(-3/2)=2/3larrcolor(blue)"slope of bisector"

$\text{using "m=2/3" and "(x_1,y_1)=(3,1)" then}$

$y - 1 = \frac{2}{3} \left(x - 3\right) \leftarrow \textcolor{red}{\text{in point-slope form}}$

$\Rightarrow y - 1 = \frac{2}{3} x - 2$

$\Rightarrow y = \frac{2}{3} x - 1 \leftarrow \textcolor{red}{\text{in slope-intercept form}}$